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# Solve the nonlinear inequality.

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Solve the nonlinear inequality.

x2 > 2(x + 4)

sally1  Jun 26, 2014

#2
+91448
+5

x2 > 2(x + 4)

You can also do a question like this graphically - which is what I would normally do.

\$\$x^2>2x+8\\\\
x^2-2x-8>0\\\\
\mbox{You can think of it like this. }\\\\
x^2-2x-8=y \mbox{ Where y is positive}\\\\
y=(x-4)(x+2)\mbox{ Where y is positive}\\\\\$\$

y=0 when x-4=0 that is when x=4  and also whatn x+2=0  that is when x=-2

So the two roots are x=4 and x=-2.

This is a concave up parabola because the number in front of the x2 is +1  (the + indicates concave up)

Do a very quick sketch - you don't need a vertex or any other features.

I'll do a good sketch because with the computer graphing programs it is too difficult to do a rough one.

You can see from the diagram that y is postive when x>4 and when x<-2

Melody  Jun 27, 2014
Sort:

#1
+576
+5

Let's move everything to the left side and distribute the two

x^2 -2x-8>0

(x-4)(x+2)>0

We can see that the two zeroes are at x=4 and x=-2, neither of which are solutions!  They are important because we can use them to determine what segment(s) of the number line contain our solution set.  Let's try testing 3 numbers, one that is greater than our two zeroes, one that is less than our two zeroes, and one that is between our zeroes.

Greater than: I'll choose x=1000 because it will make it easy to determine if our inequality holds.  Note that when you substitute a large number like this it makes both factors of our quadratic positive, which guarantees our inequality is true.

Between: I choose x=0 because it is easy to calculate. (0-4)(0+2)>0 is equivalent to -8>0 which is clearly false.

Less than: let's pick a big negative number.  It really does not matter what it is but when we select any large negative number we get two negative facts which result in a positive produce which guarantees our inequality is satisfied

Therefore our solution set is x>4 or x<-2

jboy314  Jun 27, 2014
#2
+91448
+5

x2 > 2(x + 4)

You can also do a question like this graphically - which is what I would normally do.

\$\$x^2>2x+8\\\\
x^2-2x-8>0\\\\
\mbox{You can think of it like this. }\\\\
x^2-2x-8=y \mbox{ Where y is positive}\\\\
y=(x-4)(x+2)\mbox{ Where y is positive}\\\\\$\$

y=0 when x-4=0 that is when x=4  and also whatn x+2=0  that is when x=-2

So the two roots are x=4 and x=-2.

This is a concave up parabola because the number in front of the x2 is +1  (the + indicates concave up)

Do a very quick sketch - you don't need a vertex or any other features.

I'll do a good sketch because with the computer graphing programs it is too difficult to do a rough one.

You can see from the diagram that y is postive when x>4 and when x<-2

Melody  Jun 27, 2014

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